# How does entropy change when heat is transfer into a system reversibly with constant $T$?

My guess will be

$\Delta S_\mathrm{(system)}$ : increase [because heat is flowing into the system]

$\Delta S_\mathrm{(surrounding)}$: decrease [because heat is leaving the environment]

$\Delta S_\mathrm{(universe (sys+surr))}$ : $0$, because it is a reversible process.

Please confirm if my reasonings are correct. Thank you

• by definition reversibly does not change entropy (remember entropy difference can be zero as well) – Nikos M. Sep 30 '14 at 2:38
• what do you mean by constant T in the title ? you must edit the title ( it is : "... is transfered in a reversible process ?" ? ) – user46925 Dec 26 '15 at 22:57

Entropy of system increase because of heat addition and that of surrounding decreases $\left(-{\frac{\Delta Q}{T}}\right)$.